Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $11.3$ years; the standard deviation is $2.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living between $4.7$ and $13.5$ years.
Answer: $11.3$ $9.1$ $13.5$ $6.9$ $15.7$ $4.7$ $17.9$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $11.3$ years. We know the standard deviation is $2.2$ years, so one standard deviation below the mean is $9.1$ years and one standard deviation above the mean is $13.5$ years. Two standard deviations below the mean is $6.9$ years and two standard deviations above the mean is $15.7$ years. Three standard deviations below the mean is $4.7$ years and three standard deviations above the mean is $17.9$ years. We are interested in the probability of a lion living between $4.7$ and $13.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the lions will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the mean. The probability of a particular lion living between $4.7$ and $13.5$ years is $\color{orange}{15.85\%} + {68\%}$, or $83.85\%$.